Workshop problems solving

Photovoltaic Solar Energy Conversion (PVSEC)
‫إﻧﺘﺎج اﻟﻜﻬﺮﺑﺎء ﻣﻦ اﻟﻄﺎﻗﺔ اﻟﺸﻤﺴﻴﺔ‬

Courses on photovoltaic for Moroccan academic staff; 23-27 April, ENIM / Rabat
23 27

Workshop d
W k h and problem solving activities
bl l i ti iti
Ahmed Ennaoui
Helmholtz-Zentrum Berlin für Materialien und Energie
ennaoui@helmholtz-berlin.de
This material is intended for use in lectures, presentations and as handouts to students, it can
be provided in Powerpoint format to allow customization for the individual needs of course
instructors. Permission of the author and publisher is required for any other usage.

Sources: Stanford University/JOHN WILEY & SONS, INC., PUBLICATION
Sources: Arizona State University

Aims of Classroom Workshop

To clarify and illustrate the concepts of the theoretical courses
To sharpen the communicative skills of the students, this includes
communication in other languages than the native tongue .
To develop a global quality reflex and to accomplish a reality sense:
through critical observation, in learning the applicability and limits of
the theoretical concepts and experimental techniques
To gain experience in working in groups and communicate towards
group members.
To develop the analytical mind and the sense for synthesis
To be able to report orally and by writing

Basic Laws of Radiation/ Black Body Radiation

- The following issues should be discussed:
- A photon is determined by the magnitude and the direction of its momentum
- direction of the associated electric field: p = ħ k
- Distinguish between: wave vector k with the Boltzmann constant kB.
- 2 states for photons corresponding to the two possibilities of polarization of the electric
field perpendicular to the direction of propagation.
- N b of possible states of th photons i a volume V i given by:
Number f ibl t t f the h t in l is i b

1 V 3 β=
1
ns = 2 3
dk k BT
eβ S − 1
βε
(2π)
)
Average number of photons per unit volume having their wave vector between
k and k + dk = the n mber of states x probabilit of each state
number probability


2hc2 1
E(λ, T) = 5
λ ⎡ ⎛ hc ⎞ ⎤
⎢exp⎜
⎜ λk T ⎟ − 1⎥
⎟
⎣ ⎝ B ⎠ ⎦

c = 3 0 × 108 ms-1 ; h = 6 63 × 10-34 J ; k =1.38 × 10-23 JK-1
3.0 1 6.63 34 J.s 1 38 23 1


Total area under the
curve is the total
radiant power emitted.

This area in (W/m2)
is the power emitted
between λ1 and λ2


Total area under the
curve is the total
radiant power emitted.


[Watt] = [m] x [W/m2 . K4] x [Kelvin]


Consider the earth to be a blackbody with average surface temperature
15°C and area equal to 5.1 x 1014 m2. Find the rate at which energy is
radiated by the earth and the wavelength at which maximum power is
radiated. Compare this peak wavelength with that for a 5800 K
blackbody (the sun).
The earth radiates:

σ A T 4

E (W ) = (5 64 x 10 -8 W −2 K −4 ) (5 1 x 1014 m 2 ) [(15 + 273)K]4 = 2 0 x 1017 W
(Watt) (5.64 8
Wm (5.1 2.0 Watt

The wavelength at which the maximum power is emitted

2898 2898 2898
λmax (earth) = = = 10.1μm λ (sun) = = 0.5μ.
T(K) 288 5800

Copyrighted Material, from internet

Sun (visible)
λMAX = 0.5 μm
FT = 64 million W m-2

Earth (infrared)
λMAX = 10 μm
FT = 390 W m-2
2

Copyrighted Material, from internet

Replicate this experience in your lab.

Black Body Radiation
is the amount of incoming solar radiation per unit area that would be incident on a plane
perpendicular to the rays, at a the mean distance from the Sun to the Earth.

Solar constant
2
⎛R⎞ 4
E=σ T ⎜ ⎟
⎝D⎠
σ = 5,67 . 10-8 W/(m2.K4), T = 5800 K
R = 6 96 108 m D = 1,49.1011 m
6,96.10 m, 1 49 10
For more detail
E = 1402 W/m2 See PVSEC1:Solar flux intercepted by the Earth

2 2
4 ⎛ 6,96.10 m ⎞
8
⎛R⎞
E = σ T ⎜ ⎟ = 5,67 . 10 W/(m .K ) x (5800 K) x ⎜
4 -8 2 4
⎜ 1,49.10 11 m ⎟
⎟
⎝D⎠ ⎝ ⎠
The atmosphere will transmit a fraction (75%) of solar radiation
τ E = 0,75 E = 1052 W/m2

Altitude angle at solar noon

L
Example 1: Tilt Angle of a PV Module. Find the
P δ optimum tilt angle for a south-facing
βNoon = 90 + L- δ photovoltaic module in Rabat (latitude 34° at
L
Equation
E i
Local
solar noon on M h 1
l March 1st.
horizontal

March 1st. is the 60th. day of the year so the solar declination is:

⎡ 360 ⎤ ⎡ 360 ⎤
δ = 23.45 sin ⎢ (n − 81) ⎥ = 23.45 sin ⎢
( ) (60 − 81) ⎥ = −8.3°
( )
⎣ 365 ⎦ ⎣ 365 ⎦

The tilt angle that would make the sun’s rays perpendicular
to the module at noon would therefore be Altitude
Altit d angle
l
βnoon = 90° − L + δ = 90− 34− 8.3 = 47.7 βnoon=47.7

Tilt = 90 − β noon = 90 − 47.7 = 42.3° Tilt = 42.3°
S

Altitude angle and azimuth angle
Find altitude angle β and azimuth angle S at 3 PM solar time in Boulder, CO (L = 40˚)
Boulder 40 )
on the summer solstice:.At the solstice, we know the solar declination δ = 23.45°
⎛ 15 ° ⎞ ⎛ 15 ° ⎞
HA = ⎜ ⎟ x (hours before solarnoon) = ⎜
( ou s be o e so a oo ) ⎟ x (-3h) = -45 °
(3 ) 5
⎝ h ⎠ ⎝ h ⎠
sin β = sin(40)sin (23.45) + cos(40)cos (23.45)cos (-45) = 0.7527 β = sin -1 (0.7527) = 48.8 °
cos(23.45) sin(-45)
sin(Az) = = −0 9848
0.9848 φ S = sin -1 ( 0 9848) = −80 °
i 1 (-0.9848)
cos(48.8)
A.M.
P.M.

β = 48.8 °

φ S = −80 °

Solar angles
Sunrise and sunset can be found from a simple use of:
sin(h) = sin(L)sin( δ) + cos(L)cos( δ)cos(ω) = 0

Find the time at which sunrise (geometric and conventional) will occur in
Boston (latitude 42.3°) on July 1 (n = 182). Also find conventional sunset.

Example 1: Solar Time vs. Clock Time
Find Eastern Daylight Time for solar noon in Boston (longitude 71.1 W)
71.1°
on July 1st.
Answer: July 1st. is day number n = 182. to adjust for local time,
we obtain:
360 360
B= (n − 81) = (182 − 81) = 99.89°
364 364
E = 9.87sin2B − 7.53cosB - 1.5sinB = 9.87sin2[2 (99.89)] − 7.53cos(99.89) - 1.5sin(99.89) = -3.5
( ) ( ) ( )
For Boston at longitude 71.7° W in the Eastern Time Zone with local time meridian 75°
4min
Solar Time (ST) = Clock Time (CT) + (Local Time Meridian - Local Longitude)° + E(min)
degree
To adjust for Daylight Savings Time add 1 h, so solar noon will be at about 12:48 P.M.
CT = 12 − 4(75 - 71.1)° − ( −3.5) = 12 : 00 − 12.1min = 11 : 47.9A.M. East

Origine
For Eastern
zone

Time Belts of the U.S : Eastern Standard Time - E.S.T. is calculated to the 75th meridian west longitude. Central Standard Time - C.S.T. is calculated to the 90th
meridian west. Mountain Standard Time - M.S.T. is calculated to the 105th meridian west. Pacific Standard Time - P.S.T. is calculated to the 120th meridian
west. Alaska was standardized in 1918 on 150th Meridian west, but in actual practice, other zones are and have been in use: 120° 150°, 165°

Example 2: Solar Time vs. Clock Time
For the “Green community village” near Dubai (latitude angle = 25° N, local longitude angle = 55°12’ E,
standard time zone = UTC +4, no daylight saving ti ) on February 3 at 14.00. Determine:
t d d ti 4 d li ht i time) F b t 14 00 D t i
a. the apparent solar time.
b. solar declination and hour angle , solar altitude and solar azimuth angles.
UTC = Universal Time and GMT = Greenwich Mean Time.
Atlantic Standard Time (AST) is 4 hours behind of Coordinated Universal Time(UTC)
http://www.timeanddate.com/worldclock/search.html
http://www.timeanddate.com/worldclock/results.html?query=Morocco
a) The apparent solar time 360 360
February 3th ⇒ n = 34 B =
y (n − 81) =
( ) (34 − 81) = - 46,48°
( ) ,
364 364
ET = 9.87 sin2B − 7.53B - 1.5 sinB = 9.87 sin2(-46.48) − 7.53(-46.48) - 1.5 sin(-46.48) = -13.95 min
Local longitude angle = 55°12’ E = -55.2° (conversion in °, and negative since location is in East
Standard time zone UTC+4 Local Time meridian (LTM)= -60° (negative since location is in East)
( ) ( g )
No daylight saving time February LST = 14:00
4min
AST = LST + E(min) + (LT M - Local Longitudinal angle ) = 14 : 00 + (−13.95°) + [− 60° − (−55.2°)]. 4min = 13 : 27
degree degree
Hour angle ω and solar d li ti δ
b) H l d l declination δ.
360 °(284 + n) 360 °(284 + 34)
AST = 13:27 = 13.45 h (conversion of time in hours) δ = 23.45 ° sin = 23.45 °sin = − 16.97
365 365
ω =15° (hours from local solar noon) = 15° (ST-12)
ω = 15°.( 13.45-12) = 21.75°
sin(h) = sin(L)sin(δ) + cos(L)cos(δ) ( ) ⇒ h = sin-1 [sin(25°) i (−16 97) + cos(25°)
i (h) i (L) i ( (L) (δ)cos(ω) i i (25 ).sin( 16.97) (21 75).cos(−16 97)] = 42 98°
(25 ).cos(21.75) ( 16.97) 42.98
cosδ . sinω ⎡ cos(−16.97).sin(21.75) ⎤
sin(Az) = ⇒ Az = sin −1 ⎢ ⎥ = 28.98°
cos(h) ⎣ cos(42.98) ⎦

Attenuation and air mass

Direct B
Di t Beam Radiation at th Surface of the Earth –
R di ti t the S f f th E th
Find the direct beam solar radiation normal to the sun’s rays at solar
noon on a clear day in Atlanta (latitude 33.7°C) on May 21.
y ( ) y
May 21 is day number 141

⎡ 360 ⎤
A = 1160 + 75sin . ⎢ (n − 275) ⎥ (W/m 2 ) = 1104 W/m 2
⎣ 365 ⎦
⎡ 360 ⎤
k = 0.175 + 0.035sin ⎢ (n − 100)⎥ = 0.197
( )
⎣ 365 ⎦
Optical Depth k and the apparent Extraterrestrial Flux A.
The Sky Diffuse Factor C can be used later for diffuse radiation

Measurements for the 21st Day of Each Month after ; Source: ASHRAE (1993).

Insolation on a Collector

Use the the following equation to evaluate δ
⎡ 360 ⎤
δ = 23.45 sin ⎢ (n − 81)⎥
⎣ 365 ⎦
The altitude angle of the sun at solar noon is

The air mass ratio:

The value of clear sky beam radiation at the earth’s surface:

Direct-Beam Radiation, IBC
The translation of direct-beam radiation IB (normal to the rays) into beam
insolation striking a collector face IBC is a simple function of the angle of incidence

θ = incidence angle between a normal
to the collector face and the incoming
beam.

Panel Tilt
n

Special
S i l case of beam insolation on a horizontal surface
fb i l ti h i t l f IBH

At any particular time θ will be a function of the collector orientation
time, orientation,
the altitude and azimuth angles of the sun at any particular time

The incidence angle is given by

β: altitude
φS solar azimuth
φC P
Panel azimuth
l i h

Panel Tilt

Beam Insolation on a Collector at solar noon
in Atlanta (latitude 33.7°) on May 21 the altitude angle of the sun was found to
be 76.4° and the clear-sky beam insolation was found to be 902 W/m2.
Find the beam insolation at that time on a collector that faces 20° toward the
southeast if it is tipped up at a 52° angle.
th t i ti d t l
The cosine of the incidence angle

The beam radiation on the collector β: altitude
φS solar azimuth
φC Panel azimuth

76.4°
Solar noon ΦS = 0

collector that Panel Tilt
faces 20◦ toward 52°
the southeast East

Diffuse Radiation on a Collector - find the diffuse radiation on the same panel.
Solar noon in Atlanta on May 21 (n = 141), a
Collector faces 20° toward the southeast
Tipped up at a 52° angle.
Insolation
I l ti was f found to be 902 W/ 2.
dt b W/m
Diffuse insolation on a horizontal surface IDH is proportional to the direct beam radiation IB no
matter where in the sky the sun happens to be: IDH = C x IB where C is a sky diffuse factor.

The diffuse energy striking the collector

52°

Added to the total beam insolation of 697 W/m2, this gives a total beam of 785 W/m2.

Reflected Radiation Onto a Collector.
Reflectance of the surfaces in front of the panel is 0.2 (20%)
Solar noon in Atlanta on May 21,
Altitude angle of the sun β =76.4°,
Collector faces 20° toward the southeast
Tipped up at a 52° angle
Diffuse sky factor C is 0.121
Clear-sky beam i
Cl k b insolation i 902 W/ 2.
l i is W/m 52°

The clear-sky reflected insolation on the collector is
y

The total insolation on the collector

You can use this equation, combining the equations for the three
q , g q
components of radiation:

Diffuse radiation
Direct-beam

collector, C

Reflected radiation

Tilt angle

Insolation on a Collector using obstruction diagram

Figure below

The sun path diagram with superimposed obstructions makes it easy to
estimate periods of shading at a site.

Insolation on a Collector using obstruction diagram

Physics of semiconductor devices
Continuity equations:
y q
• Derived from Maxwell’s equations:
∂
∇ ⋅ (∇ × H ) = ∇ ⋅ J cond + ∇⋅D = 0
∂t
∇⋅ (J n + J p ) + ∂ρ = 0,
∂t
ρ = q( p − n + N D − N A )
⎧ ∂n = 1 ∇ ⋅ J + G − R
⎪ ∂t q n n n
⇒ ⎨
∂p 1
⎪ = − ∇ ⋅ J p + Gp − Rp
⎩ ∂t q
• Low-level injection (SRH lifetime dominated by the minority
carrier lifetime): n p − n p0 p − pn 0
Rn = , Rp = n
τn τp

Poisson’s equation:
q
• Derived from the Maxwell’s equations (electrostatics case):
∇ × E = 0 → E = −∇ϕ
ρ
∇ ⋅ D = ∇ ⋅ (εE) = − ε∇ ϕ → ∇ ϕ = −
2 2
ε
Quasi-Fermi levels:
• In non-equilibrium conditions, one needs to define separate
Fermi levels for n and p:

⎛ E Fn − Ei ⎞
n = ni exp⎜
⎜ k T ⎟ ⎟
⎝ B ⎠ a J n = nμ n ∇E Fn
⎛ Ei − E Fp ⎞ J p = pμ p ∇E Fp
p = ni exp⎜
⎜ k T ⎟ ⎟
⎝ B ⎠
Source: Arizona State University

Sample problems:

• Decay of the photo-excited carriers
• Steady-state injection from one side
• Surface-recombination
(1) Decay of photo-excited carriers:
Consider a sample illuminated with light source until t≤0. The
generation rate equals to G At t=0 the light source is turned off
G. off.
Calculate pn(t) for t>0 .
hf

n type
n-type sample

x=0 x

Solution
• Boundary conditions: ∂p
E = 0, n =0
• Minority hole continuity equation: ∂x
∂pn 1 pn − pn 0 pn − pn 0
= − ∇ ⋅ J p + Gp − =G− Have this set up in your lab,.All
what you need a diode,
∂t q τp τp oscilloscope, )

• General form of the solution: −t / τ p
pn (t ) = Ae +B
• Boundary conditions: pn ( t )
pn 0 + τ p G
p n ( t ≤ 0 ) = p n 0 + Gτ p
pn ( ∞ ) = p n 0 Light turned off

−t / τ p
pn 0
pn (t ) = pn0 + τ pGe
Source: Arizona State University t

( )
(2) Steady-state injection from one side:
y j
Consider a sample under constant illumination by a light sour-ce. Calculate
pn(x).

hf
n-type sample

x=0 x
Solution
• Minority carrier continuity equation:

∂pn 1 1 ∂J p pn − pn 0
= − ∇ ⋅ J p + Gp − Rp ⇒ − −
∂t q q ∂x τp
2
∂pn d pn pn − pn 0
= Dp −
∂t dx 2 τp

∂pn
p
• Steady state situation:
Steady-state =0
∂t
d 2 pn pn − pn 0 d 2 Δpn Δpn
Dp 2
− =0→ 2
− 2 = 0, Lp = τ p Dp
dx τp dx Lp
− x / Lp x / Lp
Δpn ( x ) = Ae + Be
• Boundary conditions for a long sample: Diffusion length

Δpn ( ∞) = 0 → B = 0
Δpn (0) = pn (0) − pn 0 = A pn (x )
• Final solution: p n ( 0)
− x / Lp
pn ( x ) = pn 0 + [ pn (0) − pn 0 ]e
qD p − x / Lp
J p ( x) = [ pn (0) − pn0 ]e pn 0
Lp x
Lp

(3) Surface recombination:
Consider a sample under constant illumination by a light sour-ce. There is a
finite surface recombination at x=0. Calculate pn(x).
Gp

Surface recombination:
S f bi ti n-type
n type sample

dpn
Dp = S r [ p n ( 0) − p n 0 ] x
dx x =0
x 0
x=0 x L
x=L

Solution
• Minority carrier continuity equation at steady state:
steady-state:
d 2 pn pn − p n 0 d 2 Δpn Δpn Gp
Dp − + Gp = 0 ⇒ − 2 =−
dx 2 τp dx 2
Lp Dp

Surface Recombination Velocity from Photocurrent Measurements in Journal of The Electrochemical Society, 146 (12) 4640-
4646 (1999)

• General form of the solution:
− x / Lp x / Lp
Δpn ( x ) = Ae + Be +C
B=0 (asymptotic condition)

• Boundary conditions for a long sample:
Δp n ( ∞ ) = C = τ p G p
Δp n ( 0 ) = A + C = A + τ p G p
• Use boundary condition at the surface to determine Δpn(0):
dpn τ p D pG p
Dp = S r [ pn (0) − pn0 ] → Δpn (0) =
dx x = 0 D p + Sr L p
• Final solution: ⎡ Sr τ p −x / Lp ⎤
Δpn ( x) = τ pG p ⎢1 − e ⎥
⎣ L p + Sr τ p ⎦


• Graphical representation of the solution:

Δpn ( )
(x

Sr = 0
τ pG p

Sr increasing

Sr → ∞

x


Solar cell and Modules

1:


The voltage produced by the 36-cells module


4

The NOCT is cell temperature in a module when ambient is
20°C, solar irradiation is 0.8 kW/m2, and windspeed is 1 m/s.
The following expression may be used:
g p y

PV Module Performance Data Under Standard Test
Conditions (1 kW/m2, AM 1.5, 25°C Cell Temperature)


See table

Effect of Shading
To help understand shading phenomenon consider Figure (a) in which an n-cell module
phenomenon,
with current I and output voltage V shows one cell separated from the others (shown as
the top cell, though it can be any cell in the string). The equivalent circuit of the top cell
has been drawn, while the other (n − 1) cells in the string are shown as just a module with
current I and output voltage Vn-1.

all cells are in the sun and since they are in
series,
series the same current , I flows through
each of them.

Effect of Shading
Now , the top cell in figure (b) is shaded and its current source ISC has been reduced to
zero. The voltage drop across RP as current flows through it causes the diode to be
reverse biased, so the diode current is also (essentially) zero. That means the entire
current flowing through the module must travel through both RP and RS in the shaded
cell on its way to the load. That means the top cell, instead of adding to the output
voltage, actually reduces it.
We consider the bottom n − 1 cells still have full sun and
still some how carry their original current I so they will
still produce their original voltage Vn−1.
This means that the output voltage of the entire module
p g
VSH with one cell shaded will drop to:

Effect of Shading
With all n cells in the sun and carrying I , the output voltage was V so the voltage
of the bottom n − 1 cells will be

The drop in voltage ΔV at any given current I , caused by the shaded cell, is given by

Since the parallel resistance RP is so much greater than the series resistance RS

Effect of Shading
At any given current, the I –V curve for the module with one shaded cell drops
current V
by V. The huge impact this can have is illustrated in the figure below:

Effect of Shading
Example 5 : The 36-cell PV module described in Example 3 had a parallel resistance per cell of
of RP = 6.6 Ω. In full sun and at current I = 2.14 A the output voltage was found there to be V =
19.41 V. If one cell is shaded and this current some how stays the same, then:
a. What would be the new module output voltage and power?
b. What would be the voltage drop across the shaded cell?
c. How much power would be dissipated in the shaded cell?
(a) The drop in module voltage will be

The new output voltage will be 19.41 − 14.66 = 4.75 V.
Power delivered by the module with one cell shaded would be

For
F comparison, i f ll sun the module was producing 41 5 W
i in full th d l d i 41.5 W.
(b) All of that 2.14 A of current goes through the parallel plus series resistance (0.005Ω ) of
the shaded cell, so the drop across the shaded cell will be

(normally a cell in the sun will add about 0.5 V to the module; this shaded cell subtracts over 14
V from the module).

Effect of Shading
The power dissipated in the shaded cell is voltage drop times current, which is
p p g p ,

All of that power dissipated in the shaded cell is converted to heat, which can cause a local hot
spot that may permanently damage the plastic laminates enclosing the cell.

Final remarks:
The proced res demonstrated in this e ample can be e tended to de elop I –V c r es under
procedures example extended develop V curves nder
various conditions of shading. The following figures shows such curves for the example
module under full-sun conditions and with one cell 50% shaded, one cell completely shaded,
and two cells completely shaded. Also shown on the graph is a dashed vertical line at 13 V,
p y g p
which is a typical operating voltage for a module charging a 12-V battery. The reduction in
charging current for even modest amounts of shading is severe. With just one cell shaded out
of 36 in the module, the power delivered to the battery is decreased by about two-thirds!

Effect of Shading

Effects of shading on the I –V curves for a PV module. The dashed line
shows a typical voltage that the module would operate at when charging a 12-V battery;
the impact on charging current is obviously severe.

Exercice: Spectral efficiency of photovoltaic (PV) cells

This exercise will enable you to determine the theoretical maximum spectral efficiency of a
photovoltaic cell, and to determine the effect of concentration on cell efficiency.

Exercice: Spectral efficiency of photovoltaic (PV) cells

Table 2: Fractional function F0−λΤ

Solution: Spectral efficiency of photovoltaic (PV) cells

PROBLEMS
1) For the following materials determine the maximum wavelength of solar
materials,
energy capable of creating hole-electron pairs:
a. Gallium arsenide, GaAs, band gap 1.42 eV.
b. Copper indium diselenide, CuInSe2, band gap 1.01 eV
pp g p
c. Cadmium sulfide, CdS, band gap 2.42 eV.

2) A p-n junction diode at 25°C carries a current of 100 mA when the diode
voltage is 0 5 V What is the reverse saturation current, I0?
0.5 V. current

3) For the simple equivalent circuit for a 0.005 m2 photovoltaic cell shown
below, the reverse saturation current is I0 = 10−9 A and at an insolation of
1-sun the short-circuit current is ISC = 1 A,. At 25°C, find the following:

PROBLEMS
a.
a The open-circuit voltage
open circuit voltage.
b. The load current when the output voltage is V = 0.5 V.
c. The power delivered to the load when the output voltage is 0.5 V.
d. The efficiency of the cell at V = 0.5 V.

4) The equivalent circuit for a PV cell includes a parallel resistance of RP =
10 . The cell has area 0.005 m2, reverse saturation current of I0 = 10−9 A
and at an insolation of 1-sun the short-circuit current is ISC = 1 A, At 25 C,
1 sun short circuit 25◦C,
with an output voltage of 0.5 V, find the following:

PROBLEMS
a.
a The load current
current.
b. The power delivered to the load.
c. The efficiency of the cell.

5) The following figure shows two I-V curves. One is for a PV cell with an equivalent
circuit having an infinite parallel resistance (and no series resistance).
What is the parallel resistance in the equivalent circuit of the other cell?

PROBLEMS
6) The following figure shows two I-V curves. One is for a PV cell with an
IV
equivalent circuit having no series resistance (and infinite parallel resistance).
What is the series resistance in the equivalent circuit of the other cell?

PROBLEMS
7) Estimate the cell temperature and power delivered by a 100-W PV module
with the following conditions. Assume 0.5%/°C power loss.
a. NOCT = 50°C, ambient temperature of 25°C, insolation of 1-sun.
b. NOCT = 45°C, ambient temperature of 0°C, insolation of 500 W/m2.
c. NOCT = 45°C, ambient temperature of 30°C, insolation of 800 W/m2.

8) A module with 40 cells has an idealized, rectangular I-V curve with ISC = 4 A and
VOC = 20 V. If a single cell has a parallel resistance of 5 and negligible series
resistance, draw the I-V curve if one cell is completely shaded. What current would it
deliver to a 12-V battery (vertical I-V load at 12 V)?

PROBLEMS
9) Suppose a PV module has the 1 sun I V curve shown below Within the module
1-sun I-V below.
itself, the manufacturer has provided a pair of bypass diodes to help the panel
deliver some power even when many of the cells are shaded. Each diode bypasses
half of the cells, as shown. You may consider the diodes to be “ideal;” that is, they
have no voltage drop across them when conducting.

Suppose there is enough shading on the bottom cells to cause the lower diode
to start conducting. Draw the new “shaded” I-V curve for the module.

Tracking system for photovoltaics
Tracker systems are either: 2 axis trackers which track the sun both in azimuth and
2-axis trackers,
altitude angles so the collectors are always pointing directly at the sun, or 1-axis
trackers, which track only one angle or the other.

Two-axis tracking angular relationships


A single-axis tracking for photovoltaic consists of a mount having a manually
g g p g y
adjustable tilt angle along a north-south axis, and a tracking
mechanism that rotates the collector array from east-to-west

When the tilt angle of the mount is set equal to the local latitude (called a polar
mount), not only is that an optimum angle for annual collection, but the collector
geometry and resulting insolation are fairly easy to evaluate as well.

If a polar mount rotates about its axis at the same rate as the earth turns, 15°/h, then
the centerline of the collector will always face directly into the sun.

Under these conditions the incidence angle θ between a normal to the collector and the
conditions,
sun’s rays will be equal to the solar declination δ. That makes the direct-beam insolation
on the collector just egal to IB cos δ. To evaluate diffuse and reflected radiation, we need
to know the tilt angle of the collector.
g
The axis of rotation has a fixed tilt Σ = L, unless it is solar noon, the collector itself is
cocked at an odd angle with respect to the horizontal plane. The effective tilt, which is
the angle between a normal to the collector and the horizontal plane, is given by

The beam, diff
Th b diffuse, and reflected radiation on a polar mount, one-axis tracker are given
d fl t d di ti l t i t k i
by

One-Axis, Polar Mount:

Example: Compare the 40° latitude clear-sky insolation on a collector at solar noon
40 latitude,
on the summer solstice for a two-axis tracking mount versus a single-axis polar
mount. Ignore ground reflectance.
1) Solution: For two-Axis Tracker, the beam insolation from IB = Aexp (−km), we need
) , p( ),
the air mass ratio m, the apparent extraterrestrial flux A, and the optical depth k. To
find m, we need the altitude angle of the sun. Using:

with a solstice declination of 23.45°,


You can use the above relation to find A = 1088 W/m2, k = 0.205, and C = 0.134.
The direct beam insolation on the collector is therefore

The diffuse radiation on the collector is

2) One-Axis Polar Tracker: The beam portion of insolation is given by

The diffuse portion

The total is

The two-axis tracker provides 994 W/m2, which is only 9% higher
than the single-axis mount.

Sources:
Internet Materials: Arizona State University

Solar cells: operating principles, technology, and system applications by Martin Green (You have a sample of this book)

Practical photovoltaics: electricity from solar cells by Richard J. Komp.
Solar cells: What they are and how they work ; How solar cells are made ; Solar cells and modules ; Using photovoltaics ; Batteries and
other storage systems ; New developments in photovoltaic technology ; The future of photovoltaics ; Assembling your own modules.

The Physics of Solar Cells by Jenny Nelson.
Provides
P id a comprehensive introduction to the physics of the photovoltaic cell. It is suitable for undergraduates, graduate students, and
h i i t d ti t th h i f th h t lt i ll i it bl f d d t d t t d t d
researchers new to the field. It covers: basic physics of semiconductors in photovoltaic devices; physical models of solar cell operation;
characteristics and design of common types of solar cell; and approaches to increasing solar cell efficiency. The text explains the terms
and concepts of solar cell device physics and shows the reader how to formulate and solve relevant physical problems. Exercises and
worked solutions are included.

Physics of Semiconductor Devices
Classic book has set the standard for advanced study and reference in the semiconductor device field.
Designed for graduate textbook adoptions and reference needs, this new edition includes: New devices such as three-dimensional
MOSFETs, MODFETs, resonant-tunneling diodes, semiconductor sensors, quantum-cascade lasers, single-electron transistors, real-
space transfer devices, Problem sets at the end of each chapter

Photovoltaic Design & Installation For Dummies by Ryan Mayfield.
Gives you a comprehensive overview of the history, physics, design, installation, and operation of home-scale solar-panel systems. You'll
also get an introduction to the foundational mathematic and electrical concepts you need to understand and work with photovoltaic
syste s Covers all
systems. Co e s a aspects o home-scale solar-power systems , Viable resource for p o ess o a s, stude ts, a d tec ca laymen
of o e sca e so a po e syste s ab e esou ce o professionals, students, and technical ay e
Can be used to study for the NABCEP exam (the North American Board of Certified Energy Practitioners, NABCEP)
http://www.nabcep.org/about-us

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